its center of mass and then left to itself it vibrates with a definite period.
Let m = the mass of the rigid body, —
/ = the moment of inertia of the body about a vertical axis through its center of mass, I = the length of each string, 6 = the angular displacement of
the bar, <i> = the angular displacement of
the strings, T = the tensile forces of the strings,
2 D = the distance between the strings. |
In order to obtain the torque equation suppose the weight of the suspended system to be concentrated at the ends of the small bar ab and analyze the- forces acting upon it as shown in Fig. 143. Evidently ab is acted upon by four forces, namely, the tensile forces of the strings and the two forces each of which represents half the weight of the suspended system. These forces are equivalent to a couple formed by the forces F and — F? which act at the ends of ab in a horizontal direction, and a vortical force equal to the difference between the sum of the vortical components of the tensile forces of the strings and the weight of the suspended system. The vortical force givos the suspended system a motion in the vortical direction. But both this motion and the force which produces it are vory small therefore they will be neglected.
It is evident from Fig. 143 that the torque due to the horizontal couple is